Contents CHAPTER CHAPTER 2 CHAPTER

Contents CHAPTER 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 CHAPTER 2 Introduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at...
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Contents

CHAPTER

1

1.1 1.2 1.3 1.4 1.5 1.6 1.7

CHAPTER

2

Introduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant Circular Motion A Review of Trigonometry A Thousand Points of Light Computing in Calculus

Derivatives The Derivative of a Function Powers and Polynomials The Slope and the Tangent Line Derivative of the Sine and Cosine The Product and Quotient and Power Rules Limits Continuous Functions

CHAPTER

3

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

Applications of the Derivative Linear Approximation Maximum and Minimum Problems Second Derivatives: Minimum vs. Maximum Graphs Ellipses, Parabolas, and Hyperbolas Iterations x,+ = F(x,) Newton's Method and Chaos The Mean Value Theorem and l'H8pital's Rule

,

CHAPTER

Applications of the Derivative

Chapter 2 concentrated on computing derivatives. This chapter concentrates on using them. Our computations produced dyldx for functions built from xn and sin x and cos x. Knowing the slope, and if necessary also the second derivative, we can answer the questions about y =f(x) that this subject was created for: 1. How does y change when x changes? 2. What is the maximum value of y? Or the minimum? 3. How can you tell a maximum from a minimum, using derivatives? The information in dyldx is entirely local. It tells what is happening close to the point and nowhere else. In Chapter 2, Ax and Ay went to zero. Now we want to get them back. The local information explains the larger picture, because Ay is approximately dyldx times Ax. The problem is to connect the finite to the infinitesimal-the average slope to the instantaneous slope. Those slopes are close, and occasionally they are equal. Points of equality are assured by the Mean Value Theorem-which is the local-global connection at the center of differential calculus. But we cannot predict where dyldx equals AylAx. Therefore we now find other ways to recover a function from its derivatives-or to estimate distance from velocity and acceleration. It may seem surprising that we learn about y from dyldx. All our work has been going the other way! We struggled with y to squeeze out dyldx. Now we use dyldx to study y. That's life. Perhaps it really is life, to understand one generation from later generations.

3.1 Linear Approximation The book started with a straight line f = v t . The distance is linear when the velocity is constant. As soon as v begins to change, f = v t falls apart. Which velocity do we choose, when v ( t ) is not constant? The solution is to take very short time intervals,

91

3 Applications of the Derivative in which v is nearly constant:

f Af

= vt

is completely false

= vAt

is nearly true

df = vdt

is exactly true.

For a brief moment the functionf(t) is linear-and stays near its tangent line. In Section 2.3 we found the tangent line to y =f(x). At x = a, the slope of the curve and the slope of the line are f'(a). For points on the line, start at y =f(a). Add the slope times the "increment" x - a:

Y =f(a) +f '(a)(x- a).

(1)

We write a capital Y for the line and a small y for the curve. The whole point of tangents is that they are close (provided we don't move too far from a):

That is the all- urpose linear approximation. Figure 3.1 shows the square root lo, function y = A n d its tangent line at x = a = 100. At the point y = @= the slope is 1/2& = 1/20. The table beside the figure compares y(x) with Y(x).

Fig. 3.1

Y ( x )is the linear approximation to

f i near x = a = 100.

The accuracy gets worse as x departs from 100. The tangent line leaves the curve. The arrow points to a good approximation at 102, and at 101 it would be even better. In this example Y is larger than y-the straight line is above the curve. The slope of the line stays constant, and the slope of the curve is decreasing. Such a curve will soon be called "concave downward," and its tangent lines are above it. Look again at x = 102, where the approximation is good. In Chapter 2, when we were approaching dyldx, we started with Ay/Ax: slope z

JiE-m 102- 100

'

Now that is turned around! The slope is 1/20. What we don't know is J102:

JZ

w

J-5 + (slope)(102 - 100).

(4)

You work with what you have. Earlier we didn't know dyldx, so we used (3). Now we are experts at dyldx, and we use (4). After computing y' = 1/20 once and for

3.1 Linear Approximation

&

all, the tangent line stays near for every number near 100. When that nearby number is 100 + Ax, notice the error as the approximation is squared:

The desired answer is 100 + Ax, and we are off by the last term involving AX)^. The whole point of linear approximation is to ignore every term after Ax. There is nothing magic about x = 100, except that it has a nice square root. Other points and other functions allow y x Y I would like to express this same idea in different symbols. Instead of starting from a and going to x, we start from x and go a distance Ax to x Ax. The letters are different but the mathematics is identical.

+

1 3A

At any point x, and for any smooth betion y =fo,

slope at x x I

f& + h)-Ax).

(5)

Ax

+ x)" x 1 + nx for x near zero. A second important approximation: 1/(1 + x)" x 1 - nx for x near zero.

EXAMPLE 1 An important linear approximation: (1 EXAMPLE 2

Discussion Those are really the same. By changing n to - n in Example 1, it becomes Example 2. These are linear approximations using the slopes n and - n at x = 0:

( 1 + x)" z 1 + (slope at zero) times ( x - 0) = 1 + nx.

Here is the same thing with f ( x )= xn. The basepoint in equation (6)is now 1 or x:

(1 +Ax)" x 1 + nAx

( x + Ax)" z xn + nxn-'Ax.

Better than that, here are numbers. For n = 3 and

- 1 and

100, take Ax = .01:

Actually that last number is no good. The 100th power is too much. Linear approximation gives 1 100Ax = 2, but a calculator gives (l.O1)'OO= 2.7. ... This is close to e, the all-important number in Chapter 6. The binomial formula shows why the approximation failed:

+

Linear approximation forgets the AX)^ term. For Ax = 1/100 that error is nearly 3. It is too big to overlook. The exact error is f"(c), where the Mean Value Theorem in Section 3.8 places c between x and x + Ax. You already see the point: y - Y is of order AX)^. Linear approximation, quadratic error. DIFFERENTIALS

There is one more notation for this linear approximation. It has to be presented, because it is often used. The notation is suggestive and confusing at the same time-

3 Applications of the Derivative

it keeps the same symbols dx and dy that appear in the derivative. Earlier we took great pains to emphasize that dyldx is not an ordinary fraction.7 Until this paragraph, dx and dy have had no independent meaning. Now they become separate variables, like x and y but with their own names. These quantities dx and dy are called dzrerentials. The symbols dx and dy measure changes along the tangent line. They do for the approximation Y(x) exactly what Ax and Ay did for y(x). Thus dx and Ax both measure distance across. Figure 3.2 has Ax = dx. But the change in y does not equal the change in Y. One is Ay (exact for the function). The other is dy (exact for the tangent line). The differential dy is equal to AY, the change along the tangent line. Where Ay is the true change, dy is its linear approximation (dy/dx)dx. You often see dy written as f'(x)dx.

Ay = change in y (along curve) Y

dy = change in Y (along tangent)

-Axx=a

Fig. 3.2 The linear approximation to Ay is

dy =f '(x)dx.

x+dx=x+Ax

EmMPLE 3 y = x2 has dyldx = 2x so dy = 2x dx. The table has basepoint x = 2. The prediction dy differs from the true Ay by exactly (Ax)2= .0l and .04 and .09.

The differential dy =f'(x)dx is consistent with the derivative dyldx =f'(x). We finally have dy = (dy/dx)dx, but this is not as obvious as it seems! It looks like cancellation-it is really a definition. Entirely new symbols could be used, but dx and dy have two advantages: They suggest small steps and they satisfy dy =f'(x)dx. Here are three examples and three rules: d(sin x) = cos x dx

d(cf) = c df

Science and engineering and virtually all applications of mathematics depend on linear approximation. The true function is "linearized,"using its slope v: Increasing the time by At increases the distance by x vAt Increasing the force by Af increases the deflection by x vAf Increasing the production by Ap increases its value by z vAp. +Fraction or not, it is absolutely forbidden to cancel the d's.

3.1 Linear Approximation

The goal of dynamics or statics or economics is to predict this multiplier v-the derivative that equals the slope of the tangent line. The multiplier gives a local prediction of the change in the function. The exact law is nonlinear-but Ohm's law and Hooke's law and Newton's law are linear approximations. ABSOLUTE CHANGE, RELATIVE CHANGE, PERCENTAGE CHANGE

The change Ay or Af can be measured in three ways. So can Ax:

f!

Absolute change

Ax

df

Relative change

f(4

Percentage change Relative change is often more realistic than absolute change. If we know the distance to the moon within three miles, that is more impressive than knowing our own height within one inch. Absolutely, one inch is closer than three miles. Relatively, three miles is much closer: 3 miles 1 inch < or .001%< 1.4%. 300,000 miles 70 inches EXAMPLE 4 The radius of the Earth is within 80 miles of r = 4000 miles.

(a) Find the variation dV in the volume V = jnr3, using linear approximation. (b) Compute the relative variations dr/r and dV/V and AV/K Solution The job of calculus is to produce the derivative. After dV/dr = 4nr2, its work is done. The variation in volume is dV = 4n(4000)'(80) cubic miles. A 2% relative variation in r gives a 6% relative variation in V:

Without calculus we need the exact volume at r = 4000 + 80 (also at r = 3920):

One comment on dV = 4nr2dr. This is (area of sphere) times (change in radius). It is the volume of a thin shell around the sphere. The shell is added when the radius grows by dr. The exact AV/V is 3917312/640000%, but calculus just calls it 6%.

3.4

EXERCISES

Read-through questions On the graph, a linear approximation is given by the a line. At x = a, the equation for that line is Y =f(a) + b . Near x = a = 10, the linear approximation to y = x3 is Y = 1000 + c . At x = 11 the exact value is ( 1 1)3 = ' d . The approximation is Y = e . In this case Ay = f and dy = g . If we know sin x, then to estimate sin(x + Ax) we add h .

In terms of x and Ax, linear approximation is f(x + Ax) x f ( x ) + i . The error is of order (Ax)P or ( x - a)P with p = i . The differential d y equals k times the differential r . Those movements are along the m line, where Ay is along the n .

-

Find the linear approximation Y to y =f(x) near x = a: 1 f(x) =x

+ x4, a = 0

2 Ax) = l/x, a =2

96

3 Applications of the Derivative

3 f(x) = tan x, a = n/4

4 f(x) = sin x, a = n/2

5 f(x) = x sin x, a = 2n

6 f(x) = sin2x, a = 0

Compute 7-12 within .O1 by deciding on f(x), choosing the basepoint a, and evaluating f(a) +f'(a)(x - a). A calculator shows the error. 7 (2.001)(j

8 sin(.02)

9 cos(.O3)

10 ( 1 5.99)'14

11 11.98

In 23-27 find the linear change dV in the volume or d A in the surface area.

23 d V if the sides of a cube change from 10 to 10.1 24 d A if the sides of a cube change from x to x + dx. 25 d A if the radius of a sphere changes by dr. 26 d V if a circular cylinder with r = 2 changes height from 3 to 3.05 (recall V = nr2h). 27 dV if a cylinder of height 3 changes from r = 2 to r = 1.9. Extra credit: What is d V i f r and h both change (dr and dh)?

12 sin(3.14)

Calculate the numerical error in these linear approximations and compare with +(Ax)2f "(x):

13 (1.01)3z 1 + 3(.01)

14 cos(.Ol) z 1 + 0(.01)

15 (sin .01)2z 0 + 0(.01)

16 ( 1 . 0 1 ) - ~z 1 - 3(.Ol)

28 In relativity the mass is m , / J w at velocity u. By Problem 20 this is near mo + for small v. Show that the kinetic energy fmv2 and the change in mass satisfy Einstein's equation e = (Am)c2.

Confirm the approximations 19-21 by computingf'(0):

29 Enter 1.1 on your calculator. Press the square root key 5 times (slowly). What happens each time to the number after . the decimal point? This is because JGz

19 J K z 1 - f x 20 I IJ= 21 J,."u'c+

zI

+ +x2 (use f = I 1JI-u. then put u = x2)

;$ (usef ( u ) = j = ,

then put u = r 2 )

22 Write down the differentials d f for f(x) = cos x and (x + l)/(x - 1) and (.x2+ I)'.

30 In Problem 29 the numbers you see are less than 1.05, so the 1.025, . . . . The second derivative of Jlfris linear approximation is higher than the curve. 31 Enter 0.9 on your calculator and press the square root key 4 times. Predict what will appear the fifth time and press again. You now have the root of 0.9. How many decimals agree with 1 - h ( 0 .I)?

Our goal is to learn about f(x) from dfldx. We begin with two quick questions. If dfldx is positive, what does that say about f ? If the slope is negative, how is that reflected in the function? Then the third question is the critical one: How do you identify a maximum or minimum?

Normal answer: The slope is zero.

This may be the most important application of calculus, to reach df1d.x = 0. Take the easy questions first. Suppose dfldx is positive for every x between a and b. All tangent lines slope upward. The function f(x) is increasing a s x goes from n to b.

3B If dfldx > 0 then f(x) is increasing. If dfldx < 0 then f(x) is decreasing. To define increasing and decreasing, look at any two points x < X . "Increasing" requires f(x) f ( X ) . A positive slope does not mean a positive function. The function itself can be positive or negative. EXAMPLE 1 f(x) = x2 - 2x has slope 2x - 2. This slope is positive when x > 1 and negative when x < 1. The function increases after x = 1 and decreases before x = 1.

3.2

Fig. 3.3 Slopes are

Maximum and Minimum Problems

-

+. Slope is + - + - + so f is up-down-up-down-up.

We say that without computing f ( x ) at any point! The parabola in Figure 3.3 goes down to its minimum at x = 1 and up again. EXAMPLE 2 x2 - 2x + 5 has the same slope. Its graph is shifted up by 5, a number that disappears in dfldx. All functions with slope 2x - 2 are parabolas x 2 - 2x + C, shifted up or down according to C. Some parabolas cross the x axis (those crossings are solutions to f ( x ) = 0). Other parabolas stay above the axis. The solutions to x2 - 2x + 5 = 0 are complex numbers and we don't see them. The special parabola x2 - 2x + 1 = ( x - 1)2 grazes the axis at x = 1. It has a "double zero," where f ( x ) = dfldx = 0. EXAMPLE 3 Suppose dfldx = (x- l ) ( x - 2)(x - 3)(x - 4). This slope is positive beyond x = 4 and up to x = 1 (dfldx = 24 at x = 0). And dfldx is positive again between 2 and 3. At x = 1, 2, 3,4, this slope is zero and f ( x ) changes direction. Here f ( x ) is a fifth-degree polynomial, because f ' ( x ) is fourth-degree. The graph of f goes up-down-up-down-up. It might cross the x axis five times. It must cross at least once (like this one). When complex numbers are allowed, every fifth-degree polynomial has five roots.

You may feel that "positive slope implies increasing function" is obvious-perhaps it is. But there is still something delicate. Starting from dfldx > 0 at every single point, we have to deduce f ( X ) >f ( x ) at pairs of points. That is a "local to global" question, to be handled by the Mean Value Theorem. It could also wait for the Fundamental Theorem of Calculus: The diflerence f ( X ) -f ( x ) equals the area under the graph of dfldx. That area is positive, so f ( X ) exceeds f(x). MAXIMA AND MINIMA

Which x makes f ( x ) as large as possible? Where is the smallest f(x)? Without calculus we are reduced to computing values of f ( x ) and comparing. With calculus, the information is in dfldx. Suppose the maximum or minimum is at a particular point x. It is possible that the graph has a corner-and no derivative. But ifdfldx exists, it must be zero. The tangent line is level. The parabolas in Figure 3.3 change from decreasing to increasing. The slope changes from negative to positive. At this crucial point the slope is zero.

3 Applications of the Derivative

3C Local Maximum or Minimum Suppose the maximum or minimum occurs at a point x inside an interval where f(x) and df[dx are defined. Then f '(x) = 0. The word "local" allows the possibility that in other intervals, f(x) goes higher or lower. We only look near x, and we use the definition of dfldx. Start with f(x + Ax) -f(x). If f(x) is the maximum, this difference is negative or zero. The step Ax can be forward or backward: if Ax > 0: if Ax < 0:

f(x

+ AX)-f(x) ---negative < 0 Ax

positive

df 6 0. and in the limit dx

f(x+Ax)-f(x) -df 3 0. - negative 2 0 and in the limit Ax negative dx

Both arguments apply. Both conclusions dfldx dfldx = 0.

< 0 and dfldx 2 0 are correct. Thus

Maybe Richard Feynman said it best. He showed his friends a plastic curve that was made in a special way - "no matter how you turn it, the tangent at the lowest point is horizontal." They checked it out. It was true. Surely You're Joking, Mr. Feynman! is a good book (but rough on mathematicians). EXAMPLE 3 (continued) Look back at Figure 3.3b. The points that stand out

are not the "ups" or "downs" but the "turns." Those are stationary points, where dfldx = 0. We see two maxima and two minima. None of them are absolute maxima or minima, because f(x) starts at - co and ends at + co. EXAMPLE 4 f(x) = 4x3 - 3x4 has slope 12x2 - 12x3. That derivative is zero when x2 equals x3, at the two points x = 0 and x = 1. To decide between minimum and

maximum (local or absolute), the first step is to evaluate f(x) at these stationary points. We find f(0) = 0 and f(1) = 1. Now look at large x. The function goes down to - co in both directions. (You can mentally substitute x = 1000 and x = - 1000). For large x, - 3x4 dominates 4x3. Conclusion f = 1 is an absolute maximum. f = 0 is not a maximum or minimum (local or absolute). We have to recognize this exceptional possibility, that a curve (or a car) can pause for an instant (f' = 0) and continue in the same direction. The reason is the "double zero" in 12x2 - 12x3, from its double factor x2.

absolute max

Y!h

local max

-

-3

Fig. 3.4

rough point

The graphs of 4x3 - 3x4 and x + x-'. Check rough points and endpoints.

2

3.2 Maximum and Minimum Problems

+

EXAMPLE 5 Define f(x) = x x-I for x > 0. Its derivative 1 - 1/x2 is zero at x = 1. At that point f(1) = 2 is the minimum value. Every combination like f + 3 or 4 + is larger than fmin = 2. Figure 3.4 shows that the maximum of x + x- is + oo.?

'

Important The maximum always occurs at a stationarypoint (where dfldx = 0) or a rough point (no derivative) or an endpoint of the domain. These are the three types of critical points. All maxima and minima occur at critical points! At every other point df/dx > 0 or df/dx < 0. Here is the procedure: 1. Solve df/dx = 0 to find the stationary points f(x). 2. Compute f(x) at every critical point-stationary point, rough point, endpoint. 3. Take the maximum and minimum of those critical values of f(x). EXAMPLE 6 (Absolute value f(x) = 1x1) The minimum is zero at a rough point. The maximum is at an endpoint. There are no stationary points. The derivative of y = 1x1 is never zero. Figure 3.4 shows the maximum and minimum on the interval [- 3,2]. This is typical of piecewise linear functions.

Question Could the minimum be zero when the function never reaches f(x) = O? Answer Yes, f(x) = 1/(1+ x ) approaches ~ but never reaches zero as x + oo.

+

Remark 1 x + f oo and f(x) -, oo are avoided when f is continuous on a closed interval a < x < b. Then f(x) reaches its maximum and its minimum (Extreme Value Theorem). But x -,oo and f(x) + oo are too important to rule out. You test x + ca by considering large x. You recognizef(x) + oo by going above every finite value. Remark 2 Note the difference between critical points (specified by x) and critical values (specified by f(x)). The example x + x- had the minimum point x = 1 and the minimum value f(1) = 2. MAXIMUM AND MINIMUM IN APPLICATIONS

To find a maximum or minimum, solve f'(x) = 0. The slope is zero at the top and bottom of the graph. The idea is clear-and then check rough points and endpoints. But to be honest, that is not where the problem starts. In a real 'application, the first step (often the hardest) is to choose the unknown and find the function. It is we ourselves who decide on x and f(x). The equation dfldx = 0 comes in the middle of the problem, not at the beginning. I will start on a new example, with a question instead of a function. EXAMPLE 7 Where should you get onto an expressway for minimum driving time, if the expressway speed is 60 mph and ordinary driving speed is 30 mph?

I know this problem well-it comes up every morning. The Mass Pike goes to MIT and I have to join it somewhere. There is an entrance near Route 128 and another entrance further in. I used to take the second one, now I take the first. Mathematics should decide which is faster-some mornings I think they are maxima. Most models are simplified, to focus on the key idea. We will allow the expressway to be entered at any point x (Figure 3.5). Instead of two entrances (a discrete problem) ?A good word is approach when f (x) + a.Infinity is not reached. But I still say "the maximum is XI."

3 Applications of the Derivative

we have a continuous choice (a calculus problem). The trip has two parts, at speeds 30 and 60: a distance

,/-

up to the expressway, in 4 7 T 3 3 0 hours

a distance b - x on the expressway, in (b - x)/60 hours Problem

Minimize f(x) = total time =

1 1 -Jm+ -(b 30 60

-

x).

We have the function f(x). Now comes calculus. The first term uses the power rule: The derivative of u1I2is ~ ~ ' ~ ~ d Here u / d ux=.a2 + x2 has duldx = 2x:

1 1 1 f ' ( x )= -- (a2+ x 2 )- lI2(2x)- 30 2 60 To solve f '(x) = 0 , multiply by 60 and square both sides:

(a2+ x 2 ) - 'I2(2x)= 1

gives 2x = (a2+ x2)'I2 and

4x2 = a2 + x2.

(2)

Thus 3x2 = a2. This yields two candidates, x = a/& and x = - a/&. But a negative x would mean useless driving on the expressway. In fact f' is not zero at x = - a/&. That false root entered when we squared 2x. driving time f ( s ) when h > u / f i h - .\-

t**(L enter freeway

driving time f(.r) when h < u / f i

/ f ***

f * * (\-/

f***

P

\-

'1

/o

Fig. 3.5 Join the freeway at x-minimize

* h

h

*

the driving time f (x).

I notice something surprising. The stationary point x = a/& does not depend on b. The total time includes the constant b/60, which disappeared in dfldx. Somehow b must enter the answer, and this is a warning to go carefully. The minimum might occur at a rough point or an endpoint. Those are the other critical points off, and our drawing may not be realistic. Certainly we expect x 6 b, or we are entering the expressway beyond MIT. C o n t i n ~ ewith calculus. Compute the driving time f(.u) for an entrance at

The s uare root of 4a2/3 is 2a/&. We combined 2/30 - 1/60 = 3/60 and divided by Is this stationary value f * a minimum? You must look also at endpoints:

$.

enter at s = 0 : travel time is ni30

+ hi60 =f ' * *

enter at x = h: travel time is J o L

+ h2/30= f * * * .

y

3.2

Maximum and Minimum Problems

The comparison f *

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